7.10 Solid cone

2.1 Pure translational motion
2.2 Pure rotational motion
2.3 Plane motion

3.1 Angular velocity $\omega $
3.2 Angular acceleration $\left( \alpha \right)$
3.3 Kinematics equation for rotational motion
3.4 Analogy between translational motion & rotational motion

6.1 Parallel-axis theorem
6.2 Perpendicular axis theorem

7.1 Thin rod
7.2 Rectangular lamina
7.3 Circular ring
7.4 Circular disc
7.5 Solid cylinder
7.6 Cylindrical shell
7.7 Solid sphere
7.8 Hollow sphere
7.9 Spherical shell
7.10 Solid cone
7.11 Hollow cone
7.12 Hollow hemisphere
7.13 Parallelopiped
7.14 List of moment of inertia $(I)$ and radius of gyration $(K)$ of different bodies

Consider a right circular cone of radius $R$ and height $H$.

Volume of the cone, $\left( V \right) = \frac{1}{3}\pi {R^2}H$

Mass per unit volume of cone, $\left( {{\lambda _V}} \right) = \frac{{3M}}{{\pi {R^2}H}}$

Consider an infinitesimally small section of mass in the shape of a cylinder of radius $r$ and thickness $dy$ we get,

Volume of an infinitesimally small section, $\left( {dV} \right) = \pi {r^2}dy$

Mass of the infinitesimally small section, $\left( {dm} \right) = {\lambda _V}dV = \frac{{3M}}{{\pi {R^2}H}}\pi {r^2}dy$ $$dm = \frac{{3M{r^2}dy}}{{{R^2}H}}$$

Moment of inertia of a cylinder of mass $dm$ and radius $r$ about its axis is, $$\begin{equation} \begin{aligned} d{I_{COM}} = \frac{{{r^2}dm}}{2} \\ d{I_{COM}} = \frac{{{r^2}}}{2}\left( {\frac{{3M{r^2}dy}}{{{R^2}H}}} \right) \\ d{I_{COM}} = \frac{{3M{r^4}dy}}{{2{R^2}H}} \\\end{aligned} \end{equation} $$

From the property of similar triangle we get,

$$\begin{equation} \begin{aligned} \frac{{H - y}}{H} = \frac{r}{R} \\ H - y = \frac{{Hr}}{R} \\ y = H\left( {1 - \frac{r}{R}} \right) \\\end{aligned} \end{equation} $$ Also, $$dy = - \frac{H}{R}dr$$ So, $$\begin{equation} \begin{aligned} d{I_{COM}} = \frac{{3M}}{{2{R^2}H}}{r^4}\left( { - \frac{H}{R}} \right)dr \\ \int\limits_0^{{I_{COM}}} {dI} = \frac{{ - 3M}}{{2{R^3}}}\int\limits_R^0 {{r^4}dr} \\ \left[ I \right]_0^{{I_{COM}}} = \frac{{3M}}{{2{R^3}}}\int\limits_0^R {{r^4}dr} \\ \left( {{I_{COM}} - 0} \right) = \frac{{3M}}{{2{R^3}}}\left[ {\frac{{{R^5}}}{5}} \right]_0^R \\ {I_{COM}} = \frac{{3M{R^2}}}{{10}} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} {I_{COM}} = MK_{COM}^2 \\ \frac{{3M{R^2}}}{{10}} = MK_{COM}^2 \\ {K_{COM}} = \sqrt {\frac{3}{{10}}} R \\\end{aligned} \end{equation} $$